Example Problem with Complete Solution

2D-4 : Determine Properties Using Thermodynamic Tables 8 pts
Determine the missing property values in the table below for a system that contains pure H2O.
               
 
T
(oC)
P
(kPa)
v
(m3/kg)
u
(kJ/kg)
h
(kJ/kg)
x
(kg vap/kg tot)
a.)
65
25.03
2300
b.)
244
226
               
Read : The key to this problem is to recognize that all of the variables in the table are state variables (except maybe x), or properties, and that they are all intensive properties.  It is also important to assume that either one or two phases exist.  The triple point of R-134a is not common knowledge, but it is pretty safe to assume that it does not appear in this table.  We can verify this assumption later.  Also, since we have no data availabe about solid R-134a, we can assume that we have either a subcooled liquid, a superheated vapor or an equilibrium mixture of saturated vapor and saturated liquid in the system.  Gibbs Phase Rule tells us that for a pure substance in a single phase there are 2 degrees of freedom.  If two phases are present, then there is just 1 degree of freedom.  In either case, the two values of intensive properties given in each part of this problem will be sufficient to completely determine the values of all of the other intensive properties of the system.  So, we are in good shape to move forward on solving this problem.
Given : See the table in the problem statement.
Assumption : - No solid phase exists in either of these 5 systems
Lookup Data :
Saturation Data :
P*(kPa) Tsat (oC) vsat liq (m3/kg) vsat vap (m3/kg) usat liq (kJ/kg) usat vap (kJ/kg) hsat liq (kJ/kg) hsat vap (kJ/kg)
25.03 65 0.00102 6.197 272 2463.1 272 2618.2
3344 240 0.001229 0.05977 1033.2 2604.0 1037.3 2803.8
3973 250 0.001251 0.05013 1080.4 2602.4 1085.3 2801.5
200 120.2 0.001061 0.8857 504.5 2529.5 504.7 2706.6
300 133.5 0.001073 0.6058 561.1 2543.6 561.5 2725.3
Superheated Vapor Data :
P*(kPa) T (oC) v (m3/kg) u (kJ/kg) h (kJ/kg)
200 200 1.080 2654.5 2870.5
200 250 1.199 2731.2 2971.0
400 200 0.5342 2646.8 2860.5
400 250 0.5951 2726.1 2964.2
Part a.) Given : T 65 oC
P 25.03 kPa
The first step in solving each part of this problem is to determine the state of the system.  Is it subcooled liquid, superheated vapor or a two-phase VLE mixture.
We could do this by determining the boiling point or saturation temperature at the system pressure.  But, since 25.03 kPa does not appear in the Saturation Pressure Table for water, this would require an interpolation.  It is easier to determine the saturation pressure or vapor pressure based on the system temperature because 65oC does appear in the Saturation Temperature Table and therefore does not require an interpolation.
P*(kPa) Tsat (oC) vsat liq (m3/kg) vsat vap (m3/kg) usat liq (kJ/kg) usat vap (kJ/kg) hsat liq (kJ/kg) hsat vap (kJ/kg)
25.03 65 0.00102 6.197 272 2463.1 272 2618.2
P*(65oC) 25.03 kPa
Since the actual system pressure is EQUAL TO the vapor pressure, the system is at saturation.
In order to determine the values of the other properties of the system using the following equation, we will need to know the quality, x.
equation 1 Eqn 1
We can determine x from the saturation data and the known value of u for the system using :
equation 2 Eqn 2
x 0.926
Now, we can plug x back into Eqn 1 and apply it to the unknown properties, v and h.
v 5.736 m3/kg h 2443.6 kJ/kg
Part b.) Given : T 244 oC
P 226 kPa
We again begin by determining the state of the system.  Unfortunately the system temperature is not listed in the Saturation Temperature Table and the system pressure is is not listed in the Saturation Pressure Table.  Either way we go, interpolation is required.
P*(kPa) Tsat (oC) vsat liq (m3/kg) vsat vap (m3/kg) usat liq (kJ/kg) usat vap (kJ/kg) hsat liq (kJ/kg) hsat vap (kJ/kg)
3344 240 0.001229 0.05977 1033.2 2604 1037.3 2803.8
3973 250 0.001251 0.05013 1080.4 2602.4 1085.3 2801.5
200 120.2 0.001061 0.8857 504.5 2529.5 504.7 2706.6
300 133.5 0.001073 0.6058 561.1 2543.6 561.5 2725.3
We could interpolate to determine the saturation properties at 226 kPa, but there isn't much point !  Since the system temperature is higher than the saturation temperature at EITHER 200 kPa or 300 kPa, the system temperature must also be higher than the interpolated value of Tsat(226 kPa).
Since the system temperature is greater than the saturation temperature at the system pressure, the system contains a superheated vapor.  Therefore, we must use data from the Superheated Vapor Table to determine the unknown properties of the system.
x N/A - Superheated
The Superheated Vapor Table includes tables for pressure of 200 and 500 kPa, but not 226 kPa.  These two tables include rows for 200oC and 250oC, but not for 244oC.  Consequently a double interpolation is required for each unknown system propert, v, u and h.
The double interpolation can be done with the aid of tables like the ones developed in Thermo-CD.
The data required for the double interpolation tables are :
P*(kPa) T (oC) v (m3/kg) u (kJ/kg) h (kJ/kg)
200 200 1.080 2654.5 2870.5
200 250 1.199 2731.2 2971.0
400 200 0.5342 2646.8 2860.5
400 250 0.5951 2726.1 2964.2
Here is the double interpolation table for v :
Pressure (kPa)
T( oC ) 200 226 400
200 1.080 1.009 0.5342
244 1.185 1.107 0.5878
250 1.199 1.120 0.5951
I chose to interpolate on pressure first and then to interpolate on temperature.  If you do the interpolations in the opposite order, you will get a slightly different answer.  Either method is satisfactory.
v 1.107 m3/kg
Here is the double interpolation table for u :
Pressure (kPa)
T( oC ) 200 226 400
200 2654.5 2653.50 2646.8
244 2722.00 2721.29 2716.58
250 2731.2 2730.54 2726.1
I chose to interpolate on pressure first and then to interpolate on temperature.  If you do the interpolations in the opposite order, you will get a slightly different answer.  Either method is satisfactory.
u 2721.3 kJ/kg
Here is the double interpolation table for h :
Pressure (kPa)
T( oC ) 200 226 400
200 2870.5 2869.2 2860.5
244 2958.9 2958.0 2951.8
250 2971.0 2970.1 2964.2
I chose to interpolate on pressure first and then to interpolate on temperature.  If you do the interpolations in the opposite order, you will get a slightly different answer.  Either method is satisfactory.
h 2958.0 kJ/kg